Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $V$ is a vector space over a scalar field $F$. If $\dim(V)=n$, then $|V|=|F|^n$. How can I rigorously determine the cardinality of $V$ when $V$ is infinite dimensional?

My thought was that if $\mathscr{B}$ is an ordered basis for $V$, then the cardinality of $V$ is given by the set of functions from $\mathscr{B}\to F$, by identifying elements of $V$ with their $\mathscr{B}$-coordinate vector. However, I feel that we should only count functions with finite support since infinite sums don't make sense.

Is this correct? If so, how does one find the cardinality of $\{f\colon\mathscr{B}\to F\mid \mathrm{supp }(f)<\infty\}$, in terms of say $|F|$ and $|\mathscr{B}|$? Thanks.

share|improve this question
1  
Can you find the number of functions from $\mathscr{B}$ to $F$ with support of size at most $n$? –  Chris Eagle Sep 11 '12 at 18:48
    
@ChrisEagle Wouldn't that require choosing $n$ vectors in $\mathscr{B}$ to send to nonzero elements of $F$? That seems like it would already be very large since $\mathscr{B}$ is infinite. –  Nastassja Sep 11 '12 at 19:10
    
@Nastassja But what would the infinite cardinal be? –  Alex Becker Sep 11 '12 at 19:10
1  
@Nastassja The point is that the set of functions from $\mathscr B$ to $F$ with support at most $n$ is a union of at most $|\mathscr B|^n$ copies of $F^n$. This lets you calculate the cardinality using cardinal arithmetic. –  Alex Becker Sep 11 '12 at 19:19
    
@AlexBecker Thanks. May I check if I did this right? Since $B$ is infinite, $|B|^n=|B|$ for all $n$. Also, $|B||F|^n=\max\{|B|,|F|^n\}=\max\{|B|,|F|\}$ regardless of whether $F$ is finite or infinite. Doing this for all $n$, the cardinality of $V$ is $\max\{|B|,|F|\}\cdot\aleph_0=\max\{|B|,|F|\}$ anyway since $|B|\geq\aleph_0$? –  Nastassja Sep 11 '12 at 19:28

1 Answer 1

up vote 4 down vote accepted

Suppose that $V$ is a vector space over $F$ and $V$ has a basis $B$.

From the definition of a basis every $v\in V$ can be written as a unique sum of basis elements and scalars. That is, there is a finite subset of $B\times (F\setminus\{0\})$ whose sum is $v$, and if we require that this set is a function on its domain, then this set is unique.

This gives a well-defined injection from $V$ into finite subsets of $B\times(F\setminus\{0\})$. Assuming the axiom of choice we have that, $$|V|\leq\left|[B\times(F\setminus\{0\})]^{<\omega}\right|=|B\times F|=\max\{|B|,|F|\}\leq|V|\implies|V|=\max\{|B|,|F|\}.$$

share|improve this answer
    
The question asks what is the cardinality of $V$, given the dimension of $V$ and the cardinality of the scalar field. –  Chris Eagle Sep 11 '12 at 19:06
    
Thanks, but I don't see how this applies. I'm already assuming a basis is known to exist, and trying to compute the cardinality of the vector space. –  Nastassja Sep 11 '12 at 19:17
1  
Writing and revising while drinking and using iPhone keyboard is just hellish!! :-) –  Asaf Karagila Sep 11 '12 at 20:02
    
Thanks Asaf, I think this is a much neater presentation than what I said above. –  Nastassja Sep 11 '12 at 21:10
    
@Nastassja: Well to be fair I just finished my M.Sc. thesis and I had to write something like that there... :-) –  Asaf Karagila Sep 11 '12 at 21:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.